Ch.7 Equations of Motion ( 運動方程式 ) 1. 2 7.1 Equations of Motion.
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Transcript of Ch.7 Equations of Motion ( 運動方程式 ) 1. 2 7.1 Equations of Motion.
![Page 1: Ch.7 Equations of Motion ( 運動方程式 ) 1. 2 7.1 Equations of Motion.](https://reader036.fdocuments.net/reader036/viewer/2022082215/5a4d1b097f8b9ab059989747/html5/thumbnails/1.jpg)
Ch.7 Equations of Motion (Ch.7 Equations of Motion ( 運動方程式運動方程式 ))
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![Page 2: Ch.7 Equations of Motion ( 運動方程式 ) 1. 2 7.1 Equations of Motion.](https://reader036.fdocuments.net/reader036/viewer/2022082215/5a4d1b097f8b9ab059989747/html5/thumbnails/2.jpg)
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7.1 Equations of Motion
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symbol (符號 )
physical quantity (物理量 )
unit (單位 )
s displacement (位移 ) m
u initial velocity (最初速度 ) ms-1
v final velocity (最後速度 ) ms-1
t time (時間 ) s
a acceleration (加速度 ) ms-2
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Class Exercise 7.1
A ball moves down a slope with an initial velocity of 5 ms-1. The acceleration of the ball is 2 ms-2. What is the velocity of the ball 3 s later?
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Class Exercise 7.1
A ball moves down a slope with an initial velocity of 5 ms-1. The acceleration of the ball is 2 ms-2. What is the velocity of the ball 3 s later?
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v = 5 + (2)(3) = 11 ms-1
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Class Exercise 7.2
A car starting from rest accelerates at 4.0 ms-2 for 10 s. Calculate the distance travelled by the car during this time.
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Class Exercise 7.2
A car starting from rest accelerates at 4.0 ms-2 for 10 s. Calculate the distance travelled by the car during this time.
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s = (0)(10) + (0.5)(4)(10)2 = 200 m
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Class Exercise 7.3 The driver of a lorry applies brakes to stop the lorry from initial velocity of 20 ms-1. The deceleration of the lorry is 5 ms-2. What is the stopping distance of the lorry?
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Class Exercise 7.3 The driver of a lorry applies brakes to stop the lorry from initial velocity of 20 ms-1. The deceleration of the lorry is 5 ms-2. What is the stopping distance of the lorry?
200
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(0)2 = (20)2 + (2)(-5)(s) 10s = 400 s = 40 m
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Class Exercise 7.4
Peter is running with uniform acceleration along a straight road. At time t = 0, his velocity is 1.2 ms-1. After 6 s, his velocity becomes 2.4 ms-1.
(a) Calculate his acceleration.
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Class Exercise 7.4
Peter is running with uniform acceleration along a straight road. At time t = 0, his velocity is 1.2 ms-1. After 6 s, his velocity becomes 2.4 ms-1.
(a) Calculate his acceleration.
u = 1.2 ms-1
t = 6 sv = 2.4 ms-1
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Class Exercise 7.4
Peter is running with uniform acceleration along a straight road. At time t = 0, his velocity is 1.2 ms-1. After 6 s, his velocity becomes 2.4 ms-1.
(a) Calculate his acceleration.
u = 1.2 ms-1
t = 6 sv = 2.4 ms-1
v = u + at2.4 = 1.2 + (a)(6) a = 0.2 ms-2
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(b) Calculate the distance traveled in this 6 s.
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(b) Calculate the distance traveled in this 6 s.
v2 = u2 + 2as(2.4)2 = (1.2)2 + 2(0.2)(s) s = 10.8 m
OR s = ut + (0.5)at2
= (1.2)(6) + (0.5)(0.2)(6)2
s = 10.8 m
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Class Exercise 7.5
A car is moving with uniform deceleration along a straight road. Its velocity decreases from 20 ms-1 to 10 ms-1` after travelling a distance of 50 m.
(a) What is its deceleration?
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Class Exercise 7.5
A car is moving with uniform deceleration along a straight road. Its velocity decreases from 20 ms-1 to 10 ms-1` after travelling a distance of 50 m.
(a) What is its deceleration?
u = 20 ms-1
s = 50 mv = 2.4 ms-1
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Class Exercise 7.5 A car is moving with uniform deceleration along a straight road. Its velocity decreases from 20 ms-1 to 10 ms-1` after travelling a distance of 50 m.
(a) What is its deceleration?
u = 20 ms-1
s = 50 mv = 2.4 ms-1
v2 = u2 + 2as(10)2 = (20)2 + 2(a)(50) a = -3 ms-2
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(b) How much further distance will the car travel before it comes to rest?
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(b) How much further distance will the car travel before it comes to rest?
v2 = u2 + 2as(0)2 = (10)2 + 2(-3)(s) 6s = 100 s = 16.7 m
u = 10 ms-1
a = -3 ms-2
v = 0
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Homework: Assignment Book 1. Chapter 7: equations of Motion Q.1 – Q.7 (p.26-33) 2. Deadline: 2009 - ___ - ____